Thales' Theorem

Consider a right triangle $ABC,$ with $m\angle B=90^{\circ },$ inscribed in a circle.

According to the Inscribed Angle Theorem, the measure of $\angle B$ is half the measure of its intercepted arc, $\overgroup{AC}.$

Substituting $m\angle B=90^{\circ }$ into the equation above gives that $m\overgroup{AC}=180^{\circ }.$ Then, $\overgroup{AC}$ is a semicircle, implying that $\overline{AC}$ $($the hypotenuse of $△ABC)$ is a diameter of the circle.

Converse of Thales' Theorem

Consider a triangle $ABC$ inscribed in a circle such that one side of the triangle is a diameter of the circle.

Since $\overline{AC}$ is a diameter, then $\overgroup{AC}$ is a semicircle and then $m\overgroup{AC}=180^{\circ }.$ Now, the Inscribed Angle Theorem gives a relation between this arc and the angle opposite to the diameter. This implies that $\angle B$ is a right angle, which makes $△ABC$ a right triangle.